1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Laplace's Method Integration

  1. Apr 19, 2014 #1


    User Avatar
    Gold Member

    Consider the integral
    I_n(x)=\int^{2}_{1} (log_{e}t) e^{-x(t-1)^{n}}dt
    Use Laplace's Method to show that
    I_n(x) \sim \frac{1}{nx^\frac{2}{n}} \int^{\infty}_{0} \tau^{\frac{2-n}{n}} e^{-\tau} d\tau \end{equation}
    as [itex]x\rightarrow\infty[/itex].
    where [itex]0<n\leq2[/itex]. Hence find the leading order behaviour of [itex]I_{1}(x)[/itex]. and [itex]I_{2}(x)[/itex] as [itex]x\rightarrow \infty[/itex].

    Its really difficult question for me.


    [itex]g(t) = -(t-1)^{n}[/itex] has the maximum at [itex]t=0[/itex]

    but [itex]h(t)= log_{e}t[/itex] at [itex]t=0[/itex]
    so I can not go any further. PLEASE HELP ME.
  2. jcsd
  3. Apr 20, 2014 #2
    I believe you meant [itex]g[/itex] reaches its maximum at [itex]t = 1[/itex].
    Indeed, [itex]h(1) = 0[/itex], but that shouldn't be a problem, just use higher order terms in the Taylor expansion of the logarithm.
  4. Apr 20, 2014 #3


    User Avatar
    Gold Member

    at [itex] t=1 [/itex], [itex] g(t) =0 [/itex], how can i say it is maximum?

    what is the Taylor expansion of the [itex] log_{e} t[/itex]?
  5. Apr 21, 2014 #4
    [itex] g [/itex] is decreasing.

    Do you remember what a Taylor series is? That's basic if you want to calculate asymptotic expansions of integrals.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted